3
$\begingroup$

Please, I want to know different methods to prove following identity

$$\frac{\tan \theta + \sec\theta - 1}{\tan\theta-\sec\theta + 1}=\frac{1+\sin\theta}{\cos\theta}$$

$\endgroup$

closed as off-topic by Martin R, Claude Leibovici, user228113, Davide Giraudo, 3SAT Dec 30 '15 at 10:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Martin R, Claude Leibovici, Community, Davide Giraudo, 3SAT
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ It will help you and future users a great deal if you learn how to type the equations yourself instead of relying on pictures. Visit this page for a tutorial on how to type in MathJax on this site. $\endgroup$ – JMoravitz Dec 30 '15 at 9:58
  • 2
    $\begingroup$ What is the method that you already know? This helps us to give you other methods. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 30 '15 at 10:02
  • $\begingroup$ Did you try writing the LHS in terms of sin and cos and simplifying? $\endgroup$ – Gregory Grant Dec 30 '15 at 10:04
  • $\begingroup$ @Gregory Grant No I only use tan and sec as provided in question. I have not done it with sin and cos. $\endgroup$ – mnulb Dec 30 '15 at 10:06
  • $\begingroup$ RHS is $\sec \theta +\tan \theta$, may be it helps $\endgroup$ – Alex Dec 30 '15 at 10:07
3
$\begingroup$

Notice, $$LHS=\frac{\tan\theta+\sec\theta-1}{\tan\theta-\sec\theta+1}$$ $$=\frac{\frac{\sin\theta}{\cos\theta}+\frac{1}{\cos\theta}-1}{\frac{\sin\theta}{\cos\theta}-\frac{1}{\cos\theta}+1}$$ $$=\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}$$ $$=\frac{(\sin\theta-\cos\theta+1)((\sin\theta+\cos\theta)+1)}{(\sin\theta+\cos\theta-1)((\sin\theta+\cos\theta)+1)}$$ $$=\frac{\sin^2\theta+2\sin \theta+1-\cos^2\theta}{(\sin\theta+\cos\theta)^2-1}$$ $$=\frac{\sin^2\theta+2\sin \theta+\sin^2\theta}{\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta-1}$$ $$=\frac{2\sin^2\theta+2\sin \theta}{1+2\sin\theta\cos\theta-1}$$ $$=\frac{2\sin\theta(1+\sin \theta)}{2\sin\theta\cos\theta}$$ $$=\frac{1+\sin \theta}{\cos\theta}=RHS$$

$\endgroup$
1
$\begingroup$

Let's start from the complicated side.

$$\begin{align*} \frac{\tan\theta+\sec\theta-1}{\tan\theta-\sec\theta+1}&=\frac{\tan\theta+\sec\theta-1}{\tan\theta-\sec\theta+1}\times\frac{\cos\theta}{\cos\theta}\quad\text{(Multiply by $1=\frac{\cos\theta}{\cos\theta}$ to simplify)}\\ &=\frac{\sin\theta+1-\cos\theta}{\sin\theta-1+\cos\theta}\\ &=\frac{(\sin\theta+1)\left(1-\frac{\cos\theta}{\sin\theta+1}\right)}{\cos\theta\left(\frac{\sin\theta-1}{\cos\theta}+1\right)}\quad\text{(Forcefully factoring out the terms we need)}\\ &=\frac{(\sin\theta+1)\left(1-\frac{\cos\theta}{\sin\theta+1}\right)}{\cos\theta\left(\frac{\sin\theta-1}{\cos\theta}\times\frac{1+\sin\theta}{1+\sin\theta}+1\right)}\\ &=\frac{(\sin\theta+1)\left(1-\frac{\cos\theta}{\sin\theta+1}\right)}{\cos\theta\left(\frac{\sin^2\theta-1}{\cos\theta(1+\sin\theta)}+1\right)}\\ &=\frac{(\sin\theta+1)\left(1-\frac{\cos\theta}{\sin\theta+1}\right)}{\cos\theta\left(\frac{-\cos^2\theta}{\cos\theta(1+\sin\theta)}+1\right)}\\ &=\frac{(\sin\theta+1)\left(1-\frac{\cos\theta}{\sin\theta+1}\right)}{\cos\theta\left(1-\frac{\cos\theta}{1+\sin\theta}\right)}\\ &=\frac{1+\sin\theta}{\cos\theta}\quad\text{(Cancelling out common terms)} \end{align*}$$

$\endgroup$
1
$\begingroup$

$$LHS = \frac{\sin\theta+1-\cos\theta}{\sin\theta - 1+ \cos\theta}$$

$$\Longleftrightarrow\frac{\sin\theta+1-\cos\theta}{\sin\theta - 1+ \cos\theta} = \frac{1+\sin\theta}{\cos\theta}$$ $$\Longleftrightarrow(\sin\theta+1-\cos\theta)\cos\theta = (\sin\theta - 1+ \cos\theta)(1+\sin\theta)$$

$$\Longleftrightarrow\sin\theta\cos\theta+\cos\theta-\cos^2\theta = \sin\theta - 1 + \cos\theta + \sin^2\theta-\sin\theta+\cos\theta\sin\theta$$

$$\Longleftrightarrow-\cos^2\theta = -1 + \sin^2\theta$$

Done.

$\endgroup$
0
$\begingroup$

Multiplying by $\cos\theta$ the num and the denom

$$\frac{\tan \theta+\sec \theta-1}{\tan \theta-\sec \theta+1} = \frac{\sin \theta+1-\cos \theta}{\sin \theta-1+\cos \theta}$$ Now setting the equality and deleting the fractions \begin{align} \frac{\sin \theta+1-\cos \theta}{\sin \theta-1+\cos \theta}&=\frac{1+\sin \theta}{\cos \theta}\\ \cos \theta + \cos \theta - \cos^2 \theta &= \sin \theta + \sin^2 \theta-\sin \theta-1+ \cos \theta+\sin \theta \cos \theta\\ -\cos^2 \theta &=\sin^2 \theta-1. \end{align}

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.